How to find the slope of the curve at x=0 and y=0

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NITHIN XAVIER 2019-8-4

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https://ww2.mathworks.cn/matlabcentral/answers/474817-how-to-find-the-slope-of-the-curve-at-x-0-and-y-0

编辑： Star Strider 2019-8-4

I wanted to find the slope of the curve, where is cross the x-axis and y-axis (there are four points), I am a beginner in Matlab, It would be very helpful if someone can help me.

Thanks in advance.

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Walter Roberson 2019-8-4

Do you have the equations, and do you have the symbolic toolbox? If so then evaluate the derivatives at the appropriate points.

darova 2019-8-4

I want to help you

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Answer 1

Star Strider 2019-8-4

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https://ww2.mathworks.cn/matlabcentral/answers/474817-how-to-find-the-slope-of-the-curve-at-x-0-and-y-0#answer_386104

编辑：Star Strider 2019-8-4

在 MATLAB Online 中打开

Since you did not share your data, I created my own.

This should get you started:

x = linspace(-0.5, 1.5);                                                                    % Create Data
y1 = 20*exp(0.8*x)-25;                                                                      % Create Data
y2 = 15*exp(1.2*x)-35;                                                                      % Create Data
dv = [-1 0 1]*1E-7;
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                                        % Returns Approximate Zero-Crossing Indices Of Argument Vector
y1zx = zci(y1);                                                                             % Zero-Crossing Indices
dy1dx = gradient(y1,(x(2)-x(1)));                                                           % Numerical Derivative
y1y0x = interp1([y1(y1zx(1)-1),y1(y1zx(1)+1)],[x(y1zx(1)-1),x(y1zx(1)+1)],0);               % Precise Zero-Crossing
dy1y0 = interp1([x(y1zx(1)-1),x(y1zx(1)+1)],[dy1dx(y1zx(1)-1),dy1dx(y1zx(1)+1)],y1y0x);     % ‘y1’ Slope At y=0    ***
y2zx = zci(y2);                                                                             % Zero-Crossing Indices
dy2dx = gradient(y2,(x(2)-x(1)));                                                           % Numerical Derivative
y2y0x = interp1([y2(y2zx(1)-1),y2(y2zx(1)+1)],[x(y2zx(1)-1),x(y2zx(1)+1)],0);               % Precise Zero-Crossing
dy2y0 = interp1([x(y2zx(1)-1),x(y2zx(1)+1)],[dy2dx(y2zx(1)-1),dy2dx(y2zx(1)+1)],y2y0x);     % ‘y2’ Slope At y=0    ***
x0 = find(x>=0, 1, 'first');
dy1x0 = interp1([x(x0-1),x(x0+1)],[dy1dx(x0-1),dy1dx(x0+1)],0);                             % ‘y1’ Slope At x=0    ***
dy2x0 = interp1([x(x0-1),x(x0+1)],[dy2dx(x0-1),dy2dx(x0+1)],0);                             % ‘y2’ Slope At x=0    ***
figure
plot(x, y1,    x, y2)
grid
txtcely0 = sprintfc('\\leftarrow Slope = %.3f', [dy1y0, dy2y0]);
txtcelx0 = sprintfc('Slope = %.3f \\rightarrow', [dy1x0, dy2x0]);
text([y1y0x, y2y0x],[0 0], txtcely0, 'HorizontalAlignment','left', 'Rotation',-45)
text([0, 0],[y1(x0) y2(x0)], txtcelx0, 'HorizontalAlignment','right')